调用 TransformProcessType() 时，应用程序菜单不显示

Question

如果您像这样调用 TransformProcessType() ：

ProcessSerialNumber psn = { 0, kCurrentProcess }; 
(void) TransformProcessType(&psn, kProcessTransformToForegroundApplication);

那么除非您在应用程序中尽早调用它（例如在 applicationWillFinishLaunching 中），否则可可应用程序菜单不会显示。

Answer 1

我向苹果公司寻求帮助，他们给了我很好的帮助。引用 ：

调用 TransformProcessType 时不显示菜单栏的原因是您的应用程序已经是活动应用程序（即 [[NSRunningApplication currentApplication] isActive] 返回 YES），并且在以下情况下显示应用程序的菜单栏应用程序已激活

这是他们的解决方法：

- (void)transformStep1 {
    for (NSRunningApplication * app in [NSRunningApplication runningApplicationsWithBundleIdentifier:@"com.apple.finder"]) {
        [app activateWithOptions:NSApplicationActivateIgnoringOtherApps];
        break;
    }
    [self performSelector:@selector(transformStep2) withObject:nil afterDelay:0.1];
}

- (void)transformStep2
{
    ProcessSerialNumber psn = { 0, kCurrentProcess }; 
    (void) TransformProcessType(&psn, kProcessTransformToForegroundApplication);

    [self performSelector:@selector(transformStep3) withObject:nil afterDelay:0.1];
}

- (void)transformStep3
{
    [[NSRunningApplication currentApplication] activateWithOptions:NSApplicationActivateIgnoringOtherApps];
}

Answer 2

这是我让它发挥作用的方法。

BOOL MakeAppForeground()
{
    BOOL bSuccess = TranformAppToState(kProcessTransformToForegroundApplication);

    if(bSuccess)
    {
         bSuccess = (SetSystemUIMode(kUIModeNormal, 0) == 0);
        [NSApp activateIgnoringOtherApps:YES];
    }

    return bSuccess;
}

BOOL MakeAppBackground()
{
    return TranformAppToState(kProcessTransformToBackgroundApplication);
}

BOOL TranformAppToState(ProcessApplicationTransformState newState)
{
    ProcessSerialNumber psn = { 0, kCurrentProcess };
    OSStatus transformStatus = TransformProcessType(&psn, newState);

    if((transformStatus != 0))
    {
        NSError *error = [NSError errorWithDomain:NSOSStatusErrorDomain code:transformStatus userInfo:nil];
        NSLog(@"TranformAppToState: Unable to transform App state. Error - %@",error);
    }

    return (transformStatus == 0);
}

Answer 3

这是我的解决方案，它是在“首选项”窗口中单击复选框时触发的。唯一的副作用是应用程序窗口的短暂闪烁：

- (void) hideDockIcon:(bool) hide
{
    ProcessSerialNumber psn = { 0, kCurrentProcess };
    ProcessApplicationTransformState state = hide ? kProcessTransformToUIElementApplication : kProcessTransformToForegroundApplication;
    OSStatus err = TransformProcessType(&psn, state);
    
    if (err) {
        NSError *error = [NSError errorWithDomain:NSOSStatusErrorDomain code:err userInfo:nil];
        NSLog(@"TransformProcessType failed: %@", error);
    }
}

- (void) toggleDockIcon
{
    [NSApp hide:self];
    dispatch_after(dispatch_time(DISPATCH_TIME_NOW, (int64_t)(0.1 * NSEC_PER_SEC)), dispatch_get_main_queue(), ^{
        [self hideDockIcon:self.preferences.hideDockIcon];
        dispatch_after(dispatch_time(DISPATCH_TIME_NOW, (int64_t)(0.1 * NSEC_PER_SEC)), dispatch_get_main_queue(), ^{
            [NSApp activateIgnoringOtherApps:YES];
        });
    });
}